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example: if i have the following set set = [ 14, 3, 5 ]. all possible combinations(subsets) can be deducted easily however i want to deduct the subsets based on a certain rule example of a rule. a subset cannot be descending (e.g: {14,3} the otherway around is acceptable (e.g: {3,14} ) another rule is subsets cannot include two adjacent numbers (e.g {3,14} is not acceptable however {5,14} is acceptable}.

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    A set is always unordered. So there is no notion of adjacent elements in it. You should correct that to list or sequence or tuple. So you want to find all non decreasing subsequences such that adjacent elements dont appear. – 2017-02-12

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Let $S=\{x_1,x_2,\dots,x_n\}$ be the set. Since $S$ is a set then $\forall i,j\in\mathbb{N}, 1\le i,j\le n$ if $i\ne j\Rightarrow x_i\ne x_j$.

Suppose we have a subset $S'\subset S$ such that $S'$ has not adjacent numbers, then $S'$ has a unique way to "put" its numbers such that $S'=\{y_1,y_2,\dots,y_k\}$ and $y_1

So we have a bijection here, if we have a subset $S'\subset S$ and it hasn't adjacent numbers then we have one valid subset.

All we need is how calculate the subset $S'\subset S$. Then if $|S'|=k$ and $|S|=n$, all we need is to peek $k$ elements from a set of $n$ and any pair of them are adjacent numbers in $S$, then $2k\le n+1$.

Let's do this:
Given a sequence $s$ of $n$ bits then $s$ define a subset $S'\subseteq S$ such that the $i$-th bit define if the $i$-th element of $S$ has been chosen to "take part" in $S'$, if $i$-th bit is turned on $0$ then $x_i\notin S'$ and otherwise $x_i\in S'$. Also, all possible subsequence $s$ of $n$ bits define all possible subset of $S$.
So we have another bijection here, $S'$ hasn't adjacent numbers if and only if $s$ hasn't $2$ adjacent bits turned on $1$.

If we want to find a valid subset $S'$ with $k$ elements then we could do this:
Let define the sequence $s'$ of $n-k$ bits turned on $0$, then we will "put" $k$ bits turned on $1$ in $s'$ and we will take care that $s'$ hasn't $2$ adjacent bits turned on $1$. So $s'$ has $n-k+1$ possible position to "put" a bit turned on $1$, so we have $\binom{n-k+1}{k}$ possible ways to "make" a "new" sequence $s'$ of $n$ bits such that $s'$ hasn't $2$ adjacent bits turned on $1$.

So we will have $\binom{n-k+1}{k},~2k\le n+1$ subsets $S'\subset S$ of length $k$ and $S'$ hasn't $2$ adjacent numbers of the set $S$. If you want to calculate the number $n$ of all possible subset $S'$ then:
$\displaystyle n=\sum^{t}_{k=1}\binom{n-k+1}{k},$ $t=\lfloor\frac{n+1}{2}\rfloor$

Note: If you include the empty subset just take $k$ from $0$ to $t$.